JPA 查询语言标准生成器

Question

我在 EJB 容器中使用 JPA 构建了一个应用程序。 这是我的代码

@PersistenceContext(unitName = "damate-pu")
private EntityManager   em;

@Override
public Workspace find(String username, String path) {
    CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
    CriteriaQuery<Workspace> criteriaQuery = criteriaBuilder.createQuery(Workspace.class);
    Root<Workspace> from = criteriaQuery.from(Workspace.class);
    Predicate condition = criteriaBuilder.equal(from.get("Username"), username);
    Predicate condition2 = criteriaBuilder.equal(from.get("Path"), path);
    Predicate condition3 = criteriaBuilder.and(condition, condition2);
    criteriaQuery.where(condition3);
    Query query = em.createQuery(criteriaQuery);

    return (Workspace) query.getSingleResult();
}

当我尝试从 Web 服务运行此方法时，我收到以下错误： java.lang.IllegalArgumentException: The attribute [Username] from the managed type....

可能是什么问题？我觉得from.get("Username") 有问题...
你怎么看？以及如何解决？

---

编辑：Workspace.java

package com.ubb.damate.model;

import java.io.Serializable;
import javax.persistence.*;
import java.util.Date;
import java.util.Set;


/**
 * The persistent class for the workspace database table.
 * 
 */
@Entity
@Table(name="workspace")
public class Workspace implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    @Column(name="WorkspaceId", unique=true, nullable=false)
    private int workspaceId;

    @Temporal( TemporalType.DATE)
    @Column(name="CreationDate", nullable=false)
    private Date creationDate;

    @Lob()
    @Column(name="Path", nullable=false)
    private String path;

    @Column(name="Username", nullable=false, length=20)
    private String username;

    //bi-directional many-to-one association to Project
    @OneToMany(mappedBy="workspace")
    private Set<Project> projects;

    public Workspace() {
    }

    public int getWorkspaceId() {
        return this.workspaceId;
    }

    public void setWorkspaceId(int workspaceId) {
        this.workspaceId = workspaceId;
    }

    public Date getCreationDate() {
        return this.creationDate;
    }

    public void setCreationDate(Date creationDate) {
        this.creationDate = creationDate;
    }

    public String getPath() {
        return this.path;
    }

    public void setPath(String path) {
        this.path = path;
    }

    public String getUsername() {
        return this.username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    public Set<Project> getProjects() {
        return this.projects;
    }

    public void setProjects(Set<Project> projects) {
        this.projects = projects;
    }
}

Answer 1

您是否尝试过使用元模型？

CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
Metamodel m = em.getMetamodel();
EntityType<Workspace> WS = m.entity(Workspace.class);
CriteriaQuery<Workspace> criteriaQuery = criteriaBuilder.createQuery(Workspace.class);
Root<Workspace> from = criteriaQuery.from(Workspace.class);
Predicate condition = criteriaBuilder.equal(from.get(WS.username), username);

http://download.oracle.com/javaee/6/tutorial/doc/gjivm.html

Answer 2

在构建条件查询（或在字符串中构建 jpql）时，您希望使用实体属性名称，而不是列名称。您的数据库列名为“用户名”，但 Workspace 对象的属性是“用户名”，没有大写 U。

Answer 3

CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
        CriteriaQuery<> criteriaQuery = criteriaBuilder
                .createQuery(Date.class);
        Root<test> root = criteriaQuery.from(test.class);

        criteriaQuery.select(criteriaBuilder.greatest(root
                .<Date> get("Starttime")));
        criteriaQuery.where(
                criteriaBuilder.equal(root.get("columnName 1"), filtervalue),
                criteriaBuilder.equal(root.get("columnName 2"), Filtervalue));

        TypedQuery<Date> query = entityManager.createQuery(criteriaQuery);
        Date date = query.getSingleResult();